chapter13_7th_solution

Prévia do material em texto

CHAPTER 13
13.1. A uniform plane wave in air, E+x1 = E
+
x10 cos(10
10t − βz) V/m, is normally-incident on a
copper surface at z = 0. What percentage of the incident power density is transmitted into
the copper? We need to find the reflection coefficient. The intrinsic impedance of copper (a
good conductor) is
ηc =
√
jωµ
σ
= (1 + j)
√
ωµ
2σ
= (1 + j)
√
1010(4π × 107)
2(5.8× 107) = (1 + j)(.0104)
Note that the accuracy here is questionable, since we know the conductivity to only two
significant figures. We nevertheless proceed: Using η0 = 376.7288 ohms, we write
Γ =
ηc − η0
ηc + η0
=
.0104− 376.7288 + j.0104
.0104 + 376.7288 + j.0104
= −.9999 + j.0001
Now |Γ|2 = .9999, and so the transmitted power fraction is 1− |Γ|2 = .0001, or about 0.01%
is transmitted.
13.2. The plane z = 0 defines the boundary between two dielectrics. For z < 0, �r1 = 5, �′′r1 = 0,
and µ1 = µ0. For z > 0, �′r2 = 3, �
′′
r2 = 0, and µ2 = µ0. Let E
+
x1 = 200 cos(ωt− 15z) V/m and
find
a) ω: We have β = ω
√
µ0�′1 = ω
√
�′r1/c = 15. So ω = 15c/
√
�′r1 = 15 × (3 × 108)/
√
5 =
2.0× 109 s−1.
b) < S+1 >: First we need η1 =
√
µ0/�′1 = η0/
√
�′r1 = 377/
√
5 = 169 ohms. Next
we apply Eq. (76), Chapter 12, to evaluate the Poynting vector (with no loss and
consequently with no phase difference between electric and magnetic fields). We find
< S+1 >= (1/2)|E1|2/η1 az = (1/2)(200)2/169az = 119az W/m2.
c) < S−1 >: First, we need to evaluate the reflection coefficient:
Γ =
η2 − η1
η2 + η1
=
η0/
√
�′r2 − η0/
√
�′r1
η0/
√
�′r2 + η0/
√
�′r1
=
√
�′r1 −
√
�′r2√
�′r1 +
√
�′r2
=
√
5−√3√
5 +
√
3
= 0.13
Then < S−1 >= −|Γ|2 < S+1 >= −(0.13)2(119)az = −2.0az W/m2.
d) < S+2 >: This will be the remaining power, propagating in the forward z direction, or
< S+2 >= 117az W/m
2.
13.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating
regions 1 and 2. If �′′1 = �
′′
2 = 0, while �
′
r1 = µ
3
r1 and �
′
r2 = µ
3
r2, find the ratio �
′
r2/�
′
r1 if 20% of
the energy in the incident wave is reflected at the boundary. There are two possible answers.
First, since |Γ|2 = .20, and since both permittivities and permeabilities are real, Γ = ±0.447.
we then set up
Γ = ±0.447 = η2 − η1
η2 + η1
=
η0
√
(µr2/�′r2)− η0
√
(µr1/�′r1)
η0
√
(µr2/�′r2) + η0
√
(µr1/�′r1)
=
√
(µr2/µ3r2)−
√
(µr1/µ3r1)√
(µr2/µ3r2) +
√
(µr1/µ3r1)
=
µr1 − µr2
µr1 + µr2
1
13.3. (continued) Therefore
µr2
µr1
=
1∓ 0.447
1± 0.447 = (0.382, 2.62) ⇒
�′r2
�′r1
=
(
µr2
µr1
)3
= (0.056, 17.9)
13.4. A 10-MHz uniform plane wave having an initial average power density of 5W/m2 is normally-
incident from free space onto the surface of a lossy material in which �′′2/�
′
2 = 0.05, �
′
r2 = 5,
and µ2 = µ0. Calculate the distance into the lossy medium at which the transmitted wave
power density is down by 10dB from the initial 5W/m2:
First, since �′′2/�
′
2 = 0.05 << 1, we recognize region 2 as a good dielectric. Its intrinsic
impedance is therefore approximated well by Eq. (62b), Chapter 12:
η2 =
√
µ0
�′2
[
1 + j
1
2
�′′2
�′2
]
=
377√
5
[1 + j0.025]
The reflection coefficient encountered by the incident wave from region 1 is therefore
Γ =
η2 − η1
η2 + η1
=
(377/
√
5)[1 + j.025]− 377
(377/
√
5)[1 + j.025] + 377
=
(1−√5) + j.025
(1 +
√
5) + j.025
= −0.383 + j0.011
The fraction of the incident power that is reflected is then |Γ|2 = 0.147, and thus the
fraction of the power that is transmitted into region 2 is 1− |Γ|2 = 0.853. Still using the
good dielectric approximation, the attenuation coefficient in region 2 is found from Eq.
(60a), Chapter 12:
α
.=
ω�′′2
2
√
µ0
�′2
= (2π × 107)(0.05× 5× 8.854× 10−12) 377
2
√
5
= 2.34× 10−2 Np/m
Now, the power that propagates into region 2 is expressed in terms of the incident power
through
< S2 > (z) = 5(1− |Γ|2)e−2αz = 5(.853)e−2(2.34×10−2)z = 0.5 W/m2
in which the last equality indicates a factor of ten reduction from the incident power, as
occurs for a 10 dB loss. Solve for z to obtain
z =
ln(8.53)
2(2.34× 10−2) = 45.8 m
13.5. The region z < 0 is characterized by �′r = µr = 1 and �
′′
r = 0. The total E field here is given
as the sum of the two uniform plane waves, Es = 150e−j10z ax + (50� 20◦)ej10z ax V/m.
a) What is the operating frequency? In free space, β = k0 = 10 = ω/c = ω/3× 108. Thus,
ω = 3× 109 s−1, or f = ω/2π = 4.7× 108 Hz.
b) Specify the intrinsic impedance of the region z > 0 that would provide the appropriate
reflected wave: Use
Γ =
Er
Einc
=
50ej20
◦
150
=
1
3
ej20
◦
= 0.31 + j0.11 =
η − η0
η + η0
2
13.5 (continued) Now
η = η0
(
1 + Γ
1− Γ
)
= 377
(
1 + 0.31 + j0.11
1− 0.31− j0.31
)
= 691 + j177 Ω
c) At what value of z (−10 cm < z < 0) is the total electric field intensity a maximum
amplitude? We found the phase of the reflection coefficient to be φ = 20◦ = .349rad, and
we use
zmax =
−φ
2β
=
−.349
20
= −0.017 m = −1.7 cm
13.6. Region 1, z < 0, and region 2, z > 0, are described by the following parameters: �′1 = 100
pF/m, µ1 = 25 µH/m, �′′1 = 0, �
′
2 = 200 pF/m, µ2 = 50 µH/m, and �
′′
2/�
′
2 = 0.5. If
E+1 = 5e
−α1z cos(4× 109t− β1z)ax V/m, find:
a) α1: As �′′1 = 0, there is no loss mechanism that is modeled (see Eq. (44), Chapter 12),
and so α1 = 0.
b) β1: Since region 1 is lossless, the phase constant for the uniform plane wave will be
β1 = ω
√
µ1�′1 = (4× 109)
√
(25× 10−6)(100× 10−12) = 200 rad/m
c) < S+1 >: To find the power density, we need the intrinsic impedance of region 1, given by
η1 =
√
µ1
�′1
=
√
25× 10−6
100× 10−12 = 500 ohms
Then the incident power density will be
< S+1 >=
1
2η1
|E1|2 az = 5
2
2(500)
az = 25az mW/m2
d) < S−1 >: To find the reflected power, we need the intrinsic impedance of region 2. This
is found using Eq. (48), Chapter 12:
η2 =
√
µ2
�′2
1√
1− j(�′′2/�′2)
=
√
50× 10−6
200× 10−12
1√
1− j0.5 = 460 + j109 ohms
Then the reflection coefficient at the 1-2 boundary is
Γ =
η2 − η1
η2 + η1
=
460 + j109− 500
460 + j109 + 500
= −0.028 + j0.117
The reflected power fraction is then |Γ|2 = 1.44× 10−2.
Therefore < S−1 >= − < S+1 > |Γ|2 = −0.36az mW/m2.
3
13.6e) < S+2 >: We first need the attenuation coefficient in region 2. This is given by Eq. (44) in
Chapter 12, which in our case becomes
α2 = ω
√
µ2�′2
2


√
1 +
(
�′′2
�′2
)2
− 1

1/2
= (4× 109)
[
(50× 10−6)(200× 10−12)
2
]1/2 [√
1 + 0.25− 1]1/2 = 97.2 Np/m
Now
< S+2 >=< S
+
1 > (1− |Γ|2) e−2α2z = 25(0.986)e−2(97.2)z az = 24.7e−194z az mW/m2
Note the approximately 1 cm penetration depth.
13.7. The semi-infinite regions z < 0 and z > 1 m are free space. For 0 < z < 1 m, �′r = 4, µr = 1,
and �′′r = 0. A uniform plane wave with ω = 4 × 108 rad/s is travelling in the az direction
toward the interface at z = 0.
a) Find the standing wave ratio in each of the three regions: First we find the phase constant
in the middle region,
β2 =
ω
√
�′r
c
=
2(4× 108)
3× 108 = 2.67 rad/m
Then, with the middle layer thickness of 1 m, β2d = 2.67 rad. Also, the intrinsic
impedance of the middle layer is η2 = η0/
√
�′r = η0/2. We now find the input impedance:
ηin = η2
[
η0 cos(β2d) + jη2 sin(β2d)
η2 cos(β2d) + jη0 sin(β2d)
]
=
377
2
[
2 cos(2.67) + j sin(2.67)
cos(2.67) + j2 sin(2.67)
]
= 231 + j141
Now, at the first interface,
Γ12 =
ηin − η0
ηin + η0
=
231 + j141− 377
231 + j141 + 377
= −.176 + j.273 = .325 � 123◦
The standing wave ratio measured in region 1 is thus
s1 =
1 + |Γ12|
1− |Γ12| =
1 + 0.325
1− 0.325 = 1.96
In region 2the standing wave ratio is found by considering the reflection coefficient for
waves incident from region 2 on the second interface:
Γ23 =
η0 − η0/2
η0 + η0/2
=
1− 1/2
1 + 1/2
=
1
3
Then
s2 =
1 + 1/3
1− 1/3 = 2
Finally, s3 = 1, since no reflected waves exist in region 3.
4
13.7b. Find the location of the maximum |E| for z < 0 that is nearest to z = 0. We note that the
phase of Γ12 is φ = 123◦ = 2.15 rad. Thus
zmax =
−φ
2β
=
−2.15
2(4/3)
= −.81 m
13.8. A wave starts at point a, propagates 100m through a lossy dielectric for which α = 0.5 Np/m,
reflects at normal incidence at a boundary at which Γ = 0.3 + j0.4, and then returns to point
a. Calculate the ratio of the final power to the incident power after this round trip: Final
power, Pf , and incident power, Pi, are related through
Pf = Pie−2αL|Γ|2e−2αL ⇒ Pf
Pi
= |0.3 + j0.4|2e−4(0.5)100 = 3.5× 10−88(!)
Try measuring that.
13.9. Region 1, z < 0, and region 2, z > 0, are both perfect dielectrics (µ = µ0, �′′ = 0). A
uniform plane wave traveling in the az direction has a radian frequency of 3× 1010 rad/s. Its
wavelengths in the two regions are λ1 = 5 cm and λ2 = 3 cm. What percentage of the energy
incident on the boundary is
a) reflected; We first note that
�′r1 =
(
2πc
λ1ω
)2
and �′r2 =
(
2πc
λ2ω
)2
Therefore �′r1/�
′
r2 = (λ2/λ1)
2. Then with µ = µ0 in both regions, we find
Γ =
η2 − η1
η2 + η1
=
η0
√
1/�′r2 − η0
√
1/�′r1
η0
√
1/�′r2 + η0
√
1/�′r1
=
√
�′r1/�
′
r2 − 1√
�′r1/�
′
r2 + 1
=
(λ2/λ1)− 1
(λ2/λ1) + 1
=
λ2 − λ1
λ2 + λ1
=
3− 5
3 + 5
= −1
4
The fraction of the incident energy that is reflected is then |Γ|2 = 1/16 = 6.25× 10−2.
b) transmitted? We use part a and find the transmitted fraction to be
1− |Γ|2 = 15/16 = 0.938.
c) What is the standing wave ratio in region 1? Use
s =
1 + |Γ|
1− |Γ| =
1 + 1/4
1− 1/4 =
5
3
= 1.67
5
13.10. In Fig. 13.1, let region 2 be free space, while µr1 = 1, �′′r1 = 0, and �
′
r1 is unknown. Find �
′
r1 if
a) the amplitude of E−1 is one-half that of E
+
1 : Since region 2 is free space, the reflection
coefficient is
Γ =
|E−1 |
|E+1 |
=
η0 − η1
η0 + η1
=
η0 − η0/
√
�′r1
η0 + η0/
√
�′r1
=
√
�′r1 − 1√
�′r1 + 1
=
1
2
⇒ �′r1 = 9
.
b) < S−1 > is one-half of < S
+
1 >: This time
|Γ|2 =
∣∣∣∣∣
√
�′r1 − 1√
�′r1 + 1
∣∣∣∣∣
2
=
1
2
⇒ �′r1 = 34
c) |E1|min is one-half |E1|max: Use
|E1|max
|E1|min = s =
1 + |Γ|
1− |Γ| = 2 ⇒ |Γ| = Γ =
1
3
=
√
�′r1 − 1√
�′r1 + 1
⇒ �′r1 = 4
13.11. A 150 MHz uniform plane wave in normally-incident from air onto a material whose intrinsic
impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of
an electric field minimum at 0.3 wavelengths in front of the interface. Determine the impedance
of the unknown material: First, the field minimum is used to find the phase of the reflection
coefficient, where
zmin = − 12β (φ + π) = −0.3λ ⇒ φ = 0.2π
where β = 2π/λ has been used. Next,
|Γ| = s− 1
s + 1
=
3− 1
3 + 1
=
1
2
So we now have
Γ = 0.5ej0.2π =
ηu − η0
ηu + η0
We solve for ηu to find
ηu = η0(1.70 + j1.33) = 641 + j501 Ω
13.12. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.
For seawater, σ = 4 S/m, and �′r = 78.
a) Determine the fractions of the incident power that are reflected and transmitted: First
we find the loss tangent:
σ
ω�′
=
4
2π(50× 106)(78)(8.854× 10−12) = 18.4
This value is sufficiently greater than 1 to enable seawater to be considered a good con-
ductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), the intrinsic
impedance is ηs =
√
πfµ/σ(1 + j), and the reflection coefficient becomes
Γ =
√
πfµ/σ (1 + j)− η0√
πfµ/σ (1 + j) + η0
6
13.12 (continued) where
√
πfµ/σ =
√
π(50× 106)(4π × 10−7)/4 = 7.0. The fraction of the power
reflected is
Pr
Pi
= |Γ|2 = [
√
πfµ/σ − η0]2 + πfµ/σ
[
√
πfµ/σ + η0]2 + πfµ/σ
=
[7.0− 377]2 + 49.0
[7.0 + 377]2 + 49.0
= 0.93
The transmitted fraction is then
Pt
Pi
= 1− |Γ|2 = 1− 0.93 = 0.07
b) Qualitatively, how will these answers change (if at all) as the frequency is increased?
Within the limits of our good conductor approximation (loss tangent greater than about
ten), the reflected power fraction, using the formula derived in part a, is found to decrease
with increasing frequency. The transmitted power fraction thus increases.
13.13. A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab
of plexiglas (�′r = 3.45, �
′′
r = 0). Calculate the fractions of the incident power that are reflected
and transmitted. Also, describe the polarizations of the reflected and transmitted waves. First,
the impedance of the plexiglas will be η = η0/
√
3.45 = 203 Ω. Then
Γ =
203− 377
203 + 377
= −0.30
The reflected power fraction is thus |Γ|2 = 0.09. The total electric field in the plane of
the interface must rotate in the same direction as the incident field, in order to continu-
ally satisfy the boundary condition of tangential electric field continuity across the interface.
Therefore, the reflected wave will have to be left circularly polarized in order to make this
happen. The transmitted power fraction is now 1− |Γ|2 = 0.91. The transmitted field will be
right circularly polarized (as the incident field) for the same reasons.
13.14. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect con-
ductor.
a) Construct the superposition of the incident and reflected waves in phasor form: Assume
positive z travel for the incident electric field. Then, with reflection coefficient, Γ = −1,
the incident and reflected fields will add to give the total field:
Etot = Ei + Er = E0(ax + jay)e−jβz − E0(ax + jay)e+jβz
= E0

(e−jβz − ejβz)︸ ︷︷ ︸
−2j sin(βz)
ax + j
(
e−jβz − ejβz)︸ ︷︷ ︸
−2j sin(βz)
ay

 = 2E0 sin(βz) [ay − jax]
b) Determine the real instantaneous form of the result of part a:
E(z, t) = Re
{
Etotejωt
}
= 2E0 sin(βz) [cos(ωt)ay + sin(ωt)ax]
c) Describe the wave that is formed: This is a standing wave exhibiting circular polarization
in time. At each location along the z axis, the field vector rotates clockwise in the xy
plane, and has amplitude (constant with time) given by 2E0 sin(βz).
7
13.15. Consider these regions in which �′′ = 0: region 1, z < 0, µ1 = 4µH/m and �′1 = 10 pF/m;
region 2, 0 < z < 6 cm, µ2 = 2µH/m, �′2 = 25 pF/m; region 3, z > 6 cm, µ3 = µ1 and �
′
3 = �
′
1.
a) What is the lowest frequency at which a uniform plane wave incident from region 1 onto
the boundary at z = 0 will have no reflection? This frequency gives the condition β2d = π,
where d = 6 cm, and β2 = ω
√
µ2�′2 Therefore
β2d = π ⇒ ω = π
(.06)
√
µ2�′2
⇒ f = 1
0.12
√
(2× 10−6)(25× 10−12) = 1.2 GHz
b) If f = 50 MHz, what will the standing wave ratio be in region 1? At the given frequency,
β2 = (2π×5×107)
√
(2× 10−6)(25× 10−12) = 2.22 rad/m. Thus β2d = 2.22(.06) = 0.133.
The intrinsic impedance of regions 1 and 3 is η1 = η3 =
√
(4× 10−6)/(10−11) = 632 Ω.
The input impedance at the first interface is now
ηin = 283
[
632 cos(.133) + j283 sin(.133)
283 cos(.133) + j632 sin(.133)
]
= 589− j138 = 605� − .23
The reflection coefficient is now
Γ =
ηin − η1
ηin + η1
=
589− j138− 632
589− j138 + 632 = .12 � − 1.7
The standing wave ratio is now
s =
1 + |Γ|
1− |Γ| =
1 + .12
1− .12 = 1.27
13.16. A uniform plane wave in air is normally-incident onto a lossless dielectric plate of thickness
λ/8, and of intrinsic impedance η = 260 Ω. Determine the standing wave ratio in front of the
plate. Also find the fraction of the incident power that is transmitted to the other side of the
plate: With the a thickness ofλ/8, we have βd = π/4, and so cos(βd) = sin(βd) = 1
√
2. The
input impedance thus becomes
ηin = 260
[
377 + j260
260 + j377
]
= 243− j92 Ω
The reflection coefficient is then
Γ =
(243− j92)− 377
(243− j92) + 377 = −0.19− j0.18 = 0.26 � − 2.4rad
Therefore
s =
1 + .26
1− .26 = 1.7 and 1− |Γ|
2 = 1− (.26)2 = 0.93
8
13.17. Repeat Problem 13.16 for the cases in which the frequency is
a) doubled: If this is true, then d = λ/4, and thus ηin = (260)2/377 = 179. The reflection
coefficient becomes
Γ =
179− 377
179 + 377
= −0.36 ⇒ s = 1 + .36
1− .36 = 2.13
Then 1− |Γ|2 = 1− (.36)2 = 0.87.
b) quadrupled: Now, d = λ/2, and so we have a half-wave section surrounded by air. Trans-
mission will be total, and so s = 1 and 1− |Γ|2 = 1.
13.18. A uniform plane wave is normally-incident onto a slab of glass (n = 1.45) whose back surface
is in contact with a perfect conductor. Determine the reflective phase shift at the front surface
of the glass if the glass thickness is: (a) λ/2; (b) λ/4; (c) λ/8.
With region 3 being a perfect conductor, η3 = 0, and Eq. (36) gives the input impedance
to the structure as ηin = jη2 tanβ
. The reflection coefficient is then
Γ =
ηin − η0
ηin + η0
=
jη2 tanβ
− η0
jη2 tanβ
 + η0
=
η22 tan
2 β
− η20 + j2η0η2 tanβ
η22 tan
2 β
 + η20
= Γr + jΓi
where the last equality occurs by multiplying the numerator and denominator of the
middle term by the complex conjugate of its denominator. The reflective phase is now
φ = tan−1
(
Γi
Γr
)
= tan−1
[
2η2η0 tanβ
η22 tan
2 β
− η20
]
= tan−1
[
(2.90) tanβ
tanβ
− 2.10
]
where η2 = η0/1.45 has been used. We can now evaluate the phase shift for the three
given cases. First, when 
 = λ/2, β
 = π, and thus φ(λ/2) = 0. Next, when 
 = λ/4,
β
 = π/2, and
φ(λ/4)→ tan−1 [2.90] = 71◦
as 
→ λ/4. Finally, when 
 = λ/8, β
 = π/4, and
φ(λ/8) = tan−1
[
2.90
1− 2.10
]
= −69.2◦ (or 291◦)
9
13.19. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, η, known
to be different from that of free space. The thickness of each slab is λ/4, where λ is the
wavelength as measured in the slab material. The slabs are to be positioned parallel to one
another, and the combination lies in the path of a uniform plane wave, normally-incident. The
slabs are to be arranged such that the air spaces between them are either zero, one-quarter
wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and air
spaces such that
a) the wave is totally transmitted through the stack: In this case, we look for a combination
of half-wave sections. Let the inter-slab distances be d1, d2, and d3 (from left to right).
Two possibilities are i.) d1 = d2 = d3 = 0, thus creating a single section of thickness λ,
or ii.) d1 = d3 = 0, d2 = λ/2, thus yielding two half-wave sections separated by a half-
wavelength.
b) the stack presents the highest reflectivity to the incident wave: The best choice here
is to make d1 = d2 = d3 = λ/4. Thus every thickness is one-quarter wavelength. The
impedances transform as follows: First, the input impedance at the front surface of the last
slab (slab 4) is ηin,1 = η2/η0. We transform this back to the back surface of slab 3, moving
through a distance of λ/4 in free space: ηin,2 = η20/ηin,1 = η
3
0/η
2. We next transform this
impedance to the front surface of slab 3, producing ηin,3 = η2/ηin,2 = η4/η30 . We continue
in this manner until reaching the front surface of slab 1, where we find ηin,7 = η8/η70 .
Assuming η < η0, the ratio ηn/ηn−10 becomes smaller as n increases (as the number of
slabs increases). The reflection coefficient for waves incident on the front slab thus gets
close to unity, and approaches 1 as the number of slabs approaches infinity.
13.20. The 50MHz plane wave of Problem 13.12 is incident onto the ocean surface at an angle to the
normal of 60◦. Determine the fractions of the incident power that are reflected and transmitted
for
a) s polarization: To review Problem 12, we first we find the loss tangent:
σ
ω�′
=
4
2π(50× 106)(78)(8.854× 10−12) = 18.4
This value is sufficiently greater than 1 to enable seawater to be considered a good con-
ductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 11), and with µ = µ0,
the intrinsic impedance is ηs =
√
πfµ/σ(1 + j) = 7.0(1 + j). Next we need the angle of
refraction, which means that we need to know the refractive index of seawater at 50MHz.
For a uniform plane wave in a good conductor, the phase constant is
β =
nsea ω
c
.=
√
πfµσ ⇒ nsea .= c
√
µσ
4πf
= 26.8
Then, using Snell’s law, the angle of refraction is found:
sin θ2 =
nsea
n1
sin θ1 = 26.8 sin(60◦) ⇒ θ2 = 1.9◦
This angle is small enough so that cos θ2
.= 1. Therefore, for s polarization,
Γs
.=
ηs2 − ηs1
ηs2 + ηs1
=
7.0(1 + j)− 377/ cos 60◦
7.0(1 + j) + 377/ cos 60◦
= −0.98 + j0.018 = 0.98 � 179◦
10
3.20a (continued) The fraction of the power reflected is now |Γs|2 = 0.96. The fraction transmitted
is then 0.04.
b) p polarization: Again, with the refracted angle close to zero, the relection coefficient for
p polarization is
Γp
.=
ηp2 − ηp1
ηp2 + ηp1
=
7.0(1 + j)− 377 cos 60◦
7.0(1 + j) + 377 cos 60◦
= −0.93 + j0.069 = 0.93 � 176◦
The fraction of the power reflected is now |Γp|2 = 0.86. The fraction transmitted is then
0.14.
13.21. A right-circularly polarized plane wave in air is incident at Brewster’s angle onto a semi-infinite
slab of plexiglas (�′r = 3.45, �
′′
r = 0, µ = µ0).
a) Determine the fractions of the incident power that are reflected and transmitted: In
plexiglas, Brewster’s angle is θB = θ1 = tan−1(�′r2/�
′
r1) = tan
−1(
√
3.45) = 61.7◦. Then
the angle of refraction is θ2 = 90◦−θB (see Example 13.9), or θ2 = 28.3◦. With incidence
at Brewster’s angle, all p-polarized power will be transmitted — only s-polarized power
will be reflected. This is found through
Γs =
η2s − η1s
η2s + η1s
=
.614η0 − 2.11η0
.614η0 + 2.11η0
= −0.549
where η1s = η1 sec θ1 = η0 sec(61.7◦) = 2.11η0,
and η2s = η2 sec θ2 = (η0/
√
3.45) sec(28.3◦) = 0.614η0. Now, the reflected power fraction
is |Γ|2 = (−.549)2 = .302. Since the wave is circularly-polarized, the s-polarized compo-
nent represents one-half the total incident wave power, and so the fraction of the total
power that is reflected is .302/2 = 0.15, or 15%. The fraction of the incident power that
is transmitted is then the remainder, or 85%.
b) Describe the polarizations of the reflected and transmitted waves: Since all the p-polarized
component is transmitted, the reflected wave will be entirely s-polarized (linear). The
transmitted wave, while having all the incident p-polarized power, will have a reduced
s-component, and so this wave will be right-elliptically polarized.
13.22. A dielectric waveguide is shown in Fig. 13.16 with refractive indices as labeled. Incident light
enters the guide at angle φ from the front surface normal as shown. Once inside, the light
totally reflects at the upper n1 − n2 interface, where n1 > n2. All subsequent reflections from
the upper an lower boundaries will be total as well, and so the light is confined to the guide.
Express, in terms of n1 and n2, the maximum value of φ such that total confinement will
occur, with n0 = 1. The quantity sinφ is known as the numerical aperture of the guide.
From the illustration we see that φ1 maximizes when θ1 is at its minimum value. This minimum
will be the critical angle for the n1 − n2 interface, where sin θc = sin θ1 = n2/n1. Let the
refracted angle to the right of the vertical interface (not shown) be φ2, where n0 sinφ1 =
n1 sinφ2. Then we see that φ2 +θ1 = 90◦, and so sin θ1 = cosφ2. Now, the numerical aperture
becomes
sinφ1max =
n1
n0
sinφ2 = n1 cos θ1 = n1
√
1− sin2 θ1 = n1
√
1− (n2/n1)2 =
√
n21 −n22
Finally, φ1max = sin−1
(√
n21 − n22
)
is the numerical aperture angle.
11
13.23. Suppose that φ1 in Fig. 13.16 is Brewster’s angle, and that θ1 is the critical angle. Find n0 in
terms of n1 and n2: With the incoming ray at Brewster’s angle, the refracted angle of this ray
(measured from the inside normal to the front surface) will be 90◦ − φ1. Therefore, φ1 = θ1,
and thus sinφ1 = sin θ1. Thus
sinφ1 =
n1√
n20 + n
2
1
= sin θ1 =
n2
n1
⇒ n0 = (n1/n2)
√
n21 − n22
Alternatively, we could have used the result of Problem 13.22, in which it was found that
sinφ1 = (1/n0)
√
n21 − n22, which we then set equal to sin θ1 = n2/n1 to get the same result.
13.24. A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism
of Fig. 13.17 is made of glass (n = 1.45), and is in air. Considering the light path shown,
determine the apex angle, α: With entrance and exit rays at Brewster’s angle (to eliminate
reflective loss), the interior ray must be horizontal, or parallel to the bottom surface of the
prism. From the geometry, the angle between the interior ray and the normal to the prism
surfaces that it intersects is α/2. Since this angle is also Brewster’s angle, we may write:
α = 2 sin−1
(
1√
1 + n2
)
= 2 sin−1
(
1√
1 + (1.45)2
)
= 1.21 rad = 69.2◦
13.25. In the Brewster prism of Fig. 13.17, determine for s-polarized light the fraction of the incident
power that is transmitted through the prism: We use Γs = (ηs2 − ηs1)/(ηs2 + ηs1), where
ηs2 =
η2
cos(θB2)
=
η2
n/
√
1 + n2
=
η0
n2
√
1 + n2
and
ηs1 =
η1
cos(θB1)
=
η1
1/
√
1 + n2
= η0
√
1 + n2
Thus, at the first interface, Γ = (1−n2)/(1+n2). At the second interface, Γ will be equal but
of opposite sign to the above value. The power transmission coefficient through each interface
is 1− |Γ|2, so that for both interfaces, we have, with n = 1.45:
Ptr
Pinc
=
(
1− |Γ|2)2 =
[
1−
(
n2 − 1
n2 + 1
)2]2
= 0.76
12
13.26. Show how a single block of glass can be used to turn a p-polarized beam of iight through 180◦,
with the light suffering, in principle, zero reflective loss. The light is incident from air, and
the returning beam (also in air) may be displaced sideways from the incident beam. Specify
all pertinent angles and use n = 1.45 for glass. More than one design is possible here.
The prism below is designed such that light enters at Brewster’s angle, and once inside, is
turned around using total reflection. Using the result of Example 13.9, we find that with
glass, θB = 55.4◦, which, by the geometry, is also the incident angle for total reflection at the
back of the prism. For this to work, the Brewster angle must be greater than or equal to the
critical angle. This is in fact the case, since θc = sin−1(n2/n1) = sin−1(1/1.45) = 43.6◦.
13.27. Using Eq. (79) in Chapter 12 as a starting point, determine the ratio of the group and phase
velocities of an electromagnetic wave in a good conductor. Assume conductivity does not vary
with frequency: In a good conductor:
β =
√
πfµσ =
√
ωµσ
2
→ dβ
dω
=
1
2
[ωµσ
2
]−1/2 µσ
2
Thus
dω
dβ
=
(
dβ
dω
)−1
= 2
√
2ω
µσ
= vg and vp =
ω
β
=
ω√
ωµσ/2
=
√
2ω
µσ
Therefore vg/vp = 2.
13
13.28. Over a small wavelength range, the refractive index of a certain material varies approximately
linearly with wavelength as n(λ) .= na + nb(λ− λa), where na, nb, and λa are constants, and
where λ is the free space wavelength.
a) Show that d/dω = −(2πc/ω2)d/dλ: With λ as the free space wavelength, we use λ =
2πc/ω, from which dλ/dω = −2πc/ω2. Then d/dω = (dλ/dω) d/dλ = −(2πc/ω2) d/dλ.
b) Using β(λ) = 2πn/λ, determine the wavelength-dependent (or independent) group delay
over a unit distance: This will be
tg =
1
vg
=
dβ
dω
=
d
dω
[
2πn(λ)
λ
]
= −2πc
ω2
d
dλ
[
2π
λ
[na + nb(λ− λa)]
]
= −2πc
ω2
[
−2π
λ2
[na + nb(λ− λa)] + 2π
λ
nb
]
= − λ
2
2πc
[
−2πna
λ2
+
2πnbλa
λ2
]
=
1
c
(na − nbλa) s/m
c) Determine β2 from your result of part b: β2 = d2β/dω2|ω0 . Since the part b result is
independent of wavelength (and of frequency), it follows that β2 = 0.
d) Discuss the implications of these results, if any, on pulse broadening: A wavelength-
independent group delay (leading to zero β2) means that there will simply be no pulse
broadening at all. All frequency components arrive simultaneously. This sort of thing
happens in most transparent materials – that is, there will be a certain wavelength, known
as the zero dispersion wavelength, around which the variation of n with λ is locally linear.
Transmitting pulses at this wavelength will result in no pulse broadening (to first order).
13.29. A T = 5 ps transform-limited pulse propagates in a dispersive channel for which β2 =
10 ps2/km. Over what distance will the pulse spread to twice its initial width? After prop-
agation, the width is T ′ =
√
T 2 + (∆τ)2 = 2T . Thus ∆τ =
√
3T , where ∆τ = β2z/T .
Therefore
β2z
T
=
√
3T or z =
√
3T 2
β2
=
√
3(5 ps)2
10 ps2/km
= 4.3 km
13.30. A T = 20 ps transform-limited pulse propagates through 10 km of a dispersive channel for
which β2 = 12 ps2/km. The pulse then propagates through a second 10 km channel for which
β2 = −12 ps2/km. Describe the pulse at the output of the second channel and give a physical
explanation for what happened.
Our theory of pulse spreading will allow for changes in β2 down the length of the channel. In
fact, we may write in general:
∆τ =
1
T
∫ L
0
β2(z) dz
Having β2 change sign at the midpoint, yields a zero ∆τ , and so the pulse emerges from the
output unchanged! Physically, the pulse acquires a positive linear chirp (frequency increases
with time over the pulse envelope) during the first half of the channel. When β2 switches sign,
the pulse begins to acquire a negative chirp in the second half, which, over an equal distance,
will completely eliminate the chirp acquired during the first half. The pulse, if originally
transform-limited at input, will emerge, again transform-limited, at its original width. More
generally, complete dispersion compensation is achieved using a two-segment channel when
β2L = −β′2L′, assuming dispersion terms of higher order than β2 do not exist.
14